# Counting total number of cells producing antibody in a 96-well plate using Poisson

I make cells produce an antibody. I then add cells to plates that contain 96 wells (random distribution). I know a cell is in a particular well because the well will be positive for the antibody.

So in my 96-well plate I have 54 wells which are antibody-positive and therefore must contain a cell.

The problem is that there is a probability that some of these positive wells will contain 2 or even 3 cells. This means that even though there are 54 positive wells there are more than 54 cells because of the wells that contain more than one cell.

What I want to know is how many cell are in my plate. More than 54 but probably not massively so.

I also want an equation which I can use for the future. I want to be able to plug it into Excel and just use it, I am not good at math.

The critical part is the "random distribution" ... you don't really specify exactly what that means, and it will certainly matter.

If (as your tag suggests) it is random in such a way that the number of cells in each well is distributed as Poisson, and the Poisson parameter is the same for each well, and we assume the number in each well is independent of the number in any other well, then we can easily estimate the Poisson parameter.

I assume (though you didn't state it) that a well that is not positive for the antibody can contain no cells.

In that case you have 96 values from a Poisson($$\lambda$$) of which 42 were 0 and 54 were $$\geq 1$$.

Now the probability that a Poisson($$\lambda$$) takes the value $$0$$ is $$\exp(-\lambda)$$ and the probability it is $$>0$$ is $$p=1-\exp(-\lambda)$$. So you have a set of well-occupancies (0 if unoccupied, 1 if occupied) that are individually Bernoulli($$p$$) and the entire collection is binomial($$p$$).

The MLE of $$p$$ is the proportion $$\hat{p}=\frac{54}{96}$$. Consequently the MLE of $$\lambda$$ is

$$\hat{\lambda}=-\log(1-\hat{p})=\log(96)-\log(42)\approx 0.8267$$.

If we condition on that estimate we could calculate an expected number of cells in total (34.72 single cells + 14.35 doubles + 3.5 triples + 0.82 quads + 0.14 quins + ... = 79.36, or the shortcut is 96 $$\times$$ 0.8267), but just as $$\lambda$$ can differ from its estimate, the actual number of cells could vary quite widely from that.

Depending on what you seek, one approach might be to form an approximate prediction interval on the number of cells (or take a Bayesian approach and form an interval that way). There's several approaches that might be applied here if that's what you seek, but you'd need to be clearer about what your needs are (and you'd also need to be clear that such intervals will depend heavily on the assumptions).

If you're satisfied with just getting an estimate from that fitted $$\hat{\lambda}$$, if $$n$$ is the total number of wells and $$n_0$$ is the total unoccupied, that estimate of the number of cells is $$n\ln(\frac{n_\,}{n_0})$$.

This will not be an unbiased estimate but if your values are typical it could work fairly well in practice -- at least on average. You can easily obtain estimates that are very far too low or high. That is, you might now and then estimate a value that's actually about 80 as being below 60 or over 100 -- this is inherent in the nature of the problem.

• I've made a correction and some additions. Commented Nov 1, 2015 at 3:50